x+y=34 =>
y=34-x
Substitute in 1:
Let {is} means "must be"
x^2+(34-x)^2 {is} min  =>
x^2+34^2-68x+x^2 {is} min  =>
2x^2-68x {is} min  =>
2(x^2 - 34x) {is} min  =>
x^2 - 34x {is} min  =>
x^2 - 2.17.x {is} min =>
(here is the thin moment...)
x^2-2.17x+17^2 {is} min =>
(x-17)^2 {is} min =>
(x-17)^2=0
x=17
y=34-17=17

I'm just guessing...
X1/2= (+-sqr(89)-3)/4?

Last edited by krassi_holmz (2005-12-30 06:07:33)

For 2) it doesn't ask you to find the values of x, just the discriminant.
The discriminant is the bit inside the square root of the quadratic equation:

In this case, the discriminant would be 3² - 4*2*(-10) = 89.
Because it is positive, this shows that both roots will be real and distinct.

thanks a lot guys....
I have one more question:
1)Explain when a parabola will have 2 real roots, 2 imaginary roots, and 2 equal real roots.Explain what characteristics the discriminant will have for the above cases
My answer is:
discriminant>0 (2 real roots)
discriminant<0(2 imaginary roots)
discriminant=0(2 equal real roots)
is it right?

2 real roots when the discriminant is positve. 2 imaginary roots when the discriminant is negative. Two equal roots when the discriminant is zero.

If you forget what the descriminant is:

You can find the roots of a quadratic equation of the form:

ax^2 + bx + c

using the quadratic formula:

(-b +-  √ b^2 - 4ac)/2a

the part of the formula inside the radical side is the descriminant.

√ b^2 - 4ac

when the descriminant  (b^2 - 4ac) is positive, it has two real roots. When it is negative, it will require the square root of a negative number and thus only has imaginary roots. When the discriminant is 0 the roots will be -b/2a + 0, and -b/2a - 0 which are obviously equal.

Nah we just pretend to be.

Not stupid. You just need practice. Math isn't hard. It just takes some getting used to.

bery good? are you being funny?